When a certain perfect square is increased by 148, the result is anoth

\(x^2 + 148 = y^2\)

(Y-X)(Y+X)=148
2×74=148
(38-36)(38+36)=148

So \(X^2=36^2\)

Ans A (as unit digit is 6)

a^2 +148= b^2
148=b^2 - a^2
148=(b - a)*(b +a)
2*74=(38-36) *(38+36)
which gives
b=38 and a=36

square of 36 is 1296

x^2 + 148 = y^2
(x+y)(x-y) = 148

148 = 2 * 74 , By looking at options all the numbers given are above 30, since 30^2 = 900, so break down 74 into sum of 30's such that the difference is 2

74 = 38+36

(38+36) (38-36) = 148

Hence answer is 36^2 = 1296 + 148 = 1444 = 38^2

i mistakenly believed that perfect square is multiplied by 148...
so..we know x^2 +148 = y^2
148 = y^2 - x^2
148 = (x+y)(y-x)
possible options:
x+y=74
y-x=2 -> x=36, y=38

or
x+y = 37
y-x=4
can't be, since we would get y a non-integer.

now..it's either 36 or 38
look at the last digits..6^2 = 6..so last digit must be 6
8^2= 4 -> so last digit must be 4. We can eliminate B, C, and E.

36*36 = 1296 - A

38*38 = 1444 - don't have such an answer.

A

It helps to know how to quickly calculate squares of numbers.

\(30^2 = 900, 31^2 = 961 ( = (30 + 1)^2 = 900 + 1 + 2*30)\)
\(40^2 = 1600, 41^2 = 1681 (= (40+1)^2 = 1600 + 1 + 2*40)\)

We know that the distance between consecutive squares increases as we go to greater numbers.
\(30^2\) to \(31^2\) is a difference of 61.
\(40^2\) to \(41^2\) is a difference of 81.

When you add 148 to a perfect square to get another in this range, it means the numbers must be 2 apart such as 33 and 35 etc. Also they must lie between 30 and 40 because twice of 61 is 122 and twice of 81 is 162. 148 lies somewhere in between.

(A) and (B) are the only two possible options.
Consider (A) - it ends in 6 so the square root ends in 6 too. You add 148 then it will end in 4. The perfect square of number ending in 8 will end in 4. So it works.
Consider (B) too - it ends in 9 so the square root ends in 3 or 7. When you add 148, it ends in 7. No perfect square ends in 7 so this is out.

Answer (A)

"But our number properties establish that (y+x) and (y−x) must be either both odd or both even"

Can someone please explain why this is the case?

If y + x is Even, it is either of these two cases:
Even + Even = Even
Odd + Odd = Even

In both cases,
Even - Even = Even
Odd - Odd = Even

So y - x will be even when y + x is even.

Same holds for odd too.

If y + x is odd, it is either of these two cases:
Even + Odd = Odd
Odd + Even = Odd

In both cases,
Even - Odd = Odd
Odd - Even = Odd

So when y + x is odd, y - x must be odd too.

Could you please explain the above?
Why does it have to be 2 digits apart? And why have you multiplied it by 2 to find the possible range?

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\(x^2\) \(+\) \(148\) = \(y^2\)

Or, \(x^2\) = \(y^2\) \(-\) \(148\)

We know , Units digit of a square number must be - { 0 , 1 , 4 , 5 , 6 , 9 }


We are asked to find the value of original number , so by rule Original Square Number + 148 must end in 0 or 1 or 4 or 5 or 6 or 9

Now, Check the options -

A. 1296

1296 + 148 = 1444

B. 1369

1369 + 148 = 1517

C. 1681

1681 + 148 = 1829

D. 1764

1764 + 148 = 1912

E. 2500

2500 + 148 = 2648

Now, you find only option (A) and (C) are possible... So we can check these individually

(A) \(\sqrt{1444}\) = 38

(C) \(\sqrt{1829}\) = 42.77


So, Correct answer must be (A)

Only numbers that end with 1,4,5,6,9 & 0 can be perfect squares

Only choices A & C qualify after adding 148.

So for 1296 & 1681, the only perfect square that I can think of is 30^2 & 40^2
30^2 = 900 & 40^2 = 1600
Start with 1296: 1296 + 148 = 1444.
1296 is greater than 30^2 and the units digit has to be 4 of the final square.
So 1444 is either 32^2 or 38^2
38^2 = (30+8)^2 = (900 + 2*30*8 + 64) = 1444

as 148 is even, assume square roots are consecutive odd or even integers
n^2-(n-2)^2=4(n-1)
4(n-1)=148
n=38
n-2=36
36^2=1296
A

Hi All,

If you don't see the 'quadratic approach' to this question, you can still solve it with a bit of 'brute force' and some Number Property knowledge. This approach won't be elegant, but if you're comfortable doing math by hand, it will get you the correct answer.

To start, we're dealing with two PERFECT SQUARES - and perfect squares can only end in certain digits. Try listing out the numbers 1-10, inclusive AND the results when you square those numbers. You'll find that the perfect squares can end in only the following digits: 0, 1, 4, 5, 6 and 9. All of the answer choices fit that limitation, BUT we're also told that adding 148 will create another perfect square.

As an example, adding 148 to a number that ends in 0 will create a new number that ends in 8.... but THAT new number won't be a perfect square (since it ends in 8). You can eliminate Answers B, D and E for that reason.

Between Answers A and C, you just have to do a little more work to prove which is the answer:

Answer A: 1296

The square root of 1296 has to be between 30 and 40 (since 30^2 = 900 and 40^2 = 1600). Since it ends in a 6, the units digit has to be either a 4 or a 6. With a little multiplication, we can figure it out...

34^2 = 1156 which is too small
36^2 = 1296 which is a MATCH

Now we just have to see what happens when we add 148 to it...

1296+148 = 1444

From our prior work, we know that the square root of 1444 has to have a units digit of either 2 or 8. 1444 is greater than 1296, so the square root of 1444 has to be greater than 36 but less than 40. Let's try 38....

38^2 = 1444 which matches perfectly with everything else that we were told, so this must be the answer.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

I solved it this way.
No perfect square ends with digit 2,3,7,8.
So if we add 148 to any number and the result is perfect square, the result should not have unit digit 2,3,7,8.
Based on this we can strike B,D and E.
Between A and C. When you add 148 to C result is 1829.
40^2 = 1600, so C must be more than 40 square. I randomly checked for square of 43 and 42.
(40+3)^2 = 1600+9+2*40*3---> 1849.
So 1829 is not a perfect square. Answer A.

Let the numbers be x and y. y^2+148=x^2. ; or x^2-y^2=148 ; (x+y)*(x-y)=74*2 ; Solve for y ; you will get y=36. y^2 is 1296 or option A.

We can let the original value (before it’s squared) = x for some positive integer x, and the new value (before it’s squared) = x + k for some positive integer k. We can create the equation:

x^2 + 148 = (x + k)^2

x^2 + 148 = x^2 + 2kx + k^2

148 = 2kx + k^2

Since both k and x are positive, we see that k^2 < 148. Thus k ≤ 12. Also, since 148 and 2kx are even, k must be even also. Thus k can only be 2, 4, 6, 8, 10 or 12. Let’s analyze each of these values until we find a suitable value for x.

If k = 2, then

148 = 2(2)x + 2^2

144 = 4x

36 = x

We see that x can be 36 and 36^2 = 1296, which is choice A.

Answer: A

Can some more similar questions be posted on this thread?

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